= Solution
{c}
The <modulus> map is continuous, the denominator $1+|z|$ is positive everywhere, and sums, products and quotients with nonzero denominator preserve <continuity>. Hence \b[$f$ is continuous throughout $\mathbb C$].
For $g$, $|g(z)|=|z|$, so $g(z)\to0=g(0)$ as $z\to0$. The chosen argument is continuous locally away from the negative real axis, making $g$ continuous there as well. At a point $-r$, $r>0$, approach along $z=r e^{i\theta}$. From $\theta\uparrow\pi$, $g(z)\to ir$; from $\theta\downarrow-\pi$, $g(z)\to-ir$. The value assigned on the cut is $ir$. Thus \b[$g$ is discontinuous at every strictly negative real point], and therefore is not a continuous map on the entire complex plane. Its radial factor is $r$, as printed, rather than the $\sqrt r$ of a square <root>.
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